Find each of the following in the x + iyform and compare a computer solution.

$\mathbf{arcsin}\left(3i/4\right)$

The given expression is $\arcsin \left(3\mathrm{i}/4\right)$.

Represent the complex number in rectangular form means writing the given complex number in the form of x + iy in which x is the real part and y is the imaginary part.

Consider the complex number $z=\arcsin \left(3\mathrm{i}/4\right)$.

Rewrite the above expression.

$\sinh \left(\mathrm{z}\right)=\frac{3\mathrm{i}}{4}$

Write the formula for $\sin \left(\theta \right)$.

$\frac{e^{\left(z\right)}-e^{(-z)}}{2i}=\frac{3i}{4}$

Put $e^{\left(zi\right)}=u$

$$\begin{gathered} u-\frac{1}{u}=-\frac{3}{2} \\ {u}^2+1.5u-1=0 \end{gathered}$$

Write the coefficient and then substitute in the formula.

$$\begin{gathered} a=1 \\ b=1.5 \\ c=-1 \end{gathered}$$

Put in the formula.

$$\begin{gathered} u=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ u=\frac{-1.5\pm \sqrt{{(1.5)}^2+4}}{2} \\ u=\frac{-1.5\pm 2.5}{2} \\ u=-\frac{3}{4}\pm \frac{5}{4} \end{gathered}$$

Convert in rectangular form.

Find the value of $z_1$.

$z_1=\ln \left(u_1\right)$

Take n = 0,1,2,3,.... for the values below.

$$\begin{gathered} zi=\ln \left(r\right)+i\left(\theta +2n\mathrm{\pi }\right) \\ zi=\ln \left(-\frac{3}{4}-\frac{5}{4}\right) \\ zi=\ln \left(2\right)+i\left(\mathrm{\pi }\pm 2\mathrm{n\pi }\right) \end{gathered}$$

Find the value of $z_1$.

$$\begin{gathered} z_1=\frac{zi}{i} \\ {z}_1=\frac{\ln \left(2\right)+i\left(\mathrm{\pi }\pm 2\mathrm{n\pi }\right)}{i} \\ {z}_1=-i\ln \left(2\right)+\left(\mathrm{\pi }\pm 2\mathrm{n\pi }\right) \end{gathered}$$

Convert in rectangular form.

$$\begin{gathered} zi=\ln \left(u_2\right) \\ zi=\ln \left(r\right)+i\left(\theta +2n\mathrm{\pi }\right) \\ zi=\ln (0.5) \\ zi=\ln (0.5)\pm 2ni\mathrm{\pi } \end{gathered}$$

Find the value of $z_2$.

$$\begin{gathered} z_2=\frac{zi}{i} \\ {z}_2=\frac{-\ln \left(2\right)\pm 2ni\mathrm{\pi }}{i} \\ {z}_2=i\ln \left(2\right)\pm 2n\mathrm{\pi } \end{gathered}$$

Hence the general solution of the given equation$(3i/4)$

$$\begin{gathered} z_1=-i\ln \left(2\right)+\left(\mathrm{\pi }\pm 2\mathrm{n\pi }\right) \\ {z}_2=i\ln \left(2\right)\pm 2n\mathrm{\pi } \end{gathered}$$